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Charra [1.4K]
1 year ago
8

One winter day, the outside temperature, y, was never below −1°. The inside of a car was always warmer than the temperature outs

ide by at least 4°. Which graph of a system of inequalities represents this scenario?

Mathematics
2 answers:
butalik [34]1 year ago
6 0

Answer:

c

Step-by-step explanation:

mr Goodwill [35]1 year ago
4 0

Answer:

It's C

Step-by-step explanation:

Your graph should be this one, with both lines solid because, the X line, or red, has to be at least -1 or higher, and the Y line, or the blue, has to be +4 or higher

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Find c, yenvelope(x,t), and ycarrier(x,t). express your answer in terms of a, k1, k2, x, t, ω1, and ω2. separate the three parts
steposvetlana [31]

Answer:

C,Y_{envelope}(x,t), Y_{carrier}(x,t)=2A, \cos ((k_1-k_2)x/2-(\omega_1 -\omega_2)t / 2 ) , \sin ((k_1+k_2)x / 2 - (\omega_1 +\omega_2)t / 2 )

Step-by-step explanation:

Given

Y_1(x,t)=A \sin(K_1x- \omega _1 t)\\\\Y_2(x,t)=A \sin(K_2x- \omega _2 t)

using a trigonometrical identity

sin p + sin q = 2 sin ( p+q/2) cos ( p-q/2)

and here the condition is

the choice is in between sinax and cosax

where a > b

so we get using above equation

C,Y_{envelope}(x,t), Y_{carrier}(x,t)=2A, \cos ((k_1-k_2)x/2-(\omega_1 -\omega_2)t / 2 ) , \sin ((k_1+k_2)x / 2 - (\omega_1 +\omega_2)t / 2 )

4 0
2 years ago
The school play started at 2:10pm and ended at 3:22pm. How long did the play last?
Anna35 [415]
2:10=>3:10, was 1 hour
3:10=> 3:22, were 12 minutes

the play last 1 hour 12 minutes (or 60+12= 72 minutes).
3 0
1 year ago
Read 2 more answers
)A mule deer can run one over four of a mile in 25 seconds. At this rate, which expression can be used to determine how fast a m
makvit [3.9K]
Well, if you take x to be 25 seconds, there are 2.4x in 1min. So, that multiplied by 60, would be 144x. Therefore, 1/4 multiplied by 144x should be the answer.
6 0
2 years ago
Read 2 more answers
A rectangular piece of land is 40m long and 25m wide. A path of uniform width and 426 m^2 area sorrounds
fredd [130]

Answer:

Width of the uniform path that surrounds the piece of land = 3 m

Step-by-step explanation:

Let the width of the path that surrounds the piece of land be x.

The situation described is sketched in the attached image to this solution.

The dimension of the Length of the piece of land including the uniform path that surrounds it = (40 + 2x) m

The Breadth of the piece of land including the uniform path that surrounds it = (25 + 2x) m

The area of the piece of land including the uniform path that surrounds it

= (Area of the piece of land) + (Area of the uniform path that surrounds it)

Area of the piece of land = Length × Breadth = 40 × 25 = 1000 m²

Area of the uniform path that surrounds it = 426 m² (Given in the question)

The area of the piece of land including the uniform path that surrounds it = 1000 + 426 = 1426 m²

But the area of the piece of land including the uniform path that surrounds it is also

= (Length of the piece of land including the uniform path that surrounds it) × (Breadth of the piece of land including the uniform path that surrounds it

= (40 + 2x) + (25 + 2x)

= 1000 + 80x + 50x + 4x²

= (4x² + 130x + 1000) m²

Equation these 2 areas

4x² + 130x + 1000 = 1426

4x² + 130x - 426 = 0

Solving the quadratic equation

x = 3 or -35.5

Since the width cannot be negative,

x = 3 m

Hope this Helps!!!

7 0
2 years ago
A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if
topjm [15]

Answer:

a) The expected value is \frac{-1}{15}

b) The variance is  \frac{49}{45}

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is \binom{10}{2}.

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is \binom{5}{2}. Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is \binom{5}{1}\cdot \binom{5}{1}. So, we define the following probabilities.

Probability of winning: \frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}

Probability of losing \frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing =1.10\cdot  \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing =1.10^2\cdot  \frac{4}{9}+\frac{5}{9}=\frac{82}{75}

We will use the following formula

Var(X) = E(X^2)-E(X)^2

Thus

Var(X) = \frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}

7 0
2 years ago
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